Problem: You have found the following ages (in years) of $5$ zebras. Those zebras were randomly selected from the $42$ zebras at your local zoo: $ 8,\enspace 11,\enspace 17,\enspace 7,\enspace 19$ Based on your sample, what is the average age of the zebras? What is the standard deviation? Round your answers to the nearest tenth. Average age: $ $
Because we only have data for a small sample of the 42 zebras, we are only able to estimate the population mean and standard deviation by finding the sample mean $({\overline{x}})$ and sample standard deviation $({s})$. To find the sample mean, add up the values of all $5$ samples and divide by $5$. $ {\overline{x}} = \dfrac{\sum\limits_{i=1}^{{n}} x_i}{{n}} = \dfrac{\sum\limits_{i=1}^{{5}} x_i}{{5}} $ $ {\overline{x}} = \dfrac{8 + 11 + 17 + 7 + 19}{{5}} = {12.4\text{ years old}} $ Find the squared deviations from the mean for each sample. Since we don't know the population mean, estimate the mean by using the sample mean we just calculated $({\overline{x}} = {12.4\text{ years}})$. Age $x_i$ Distance from the mean $(x_i - {\overline{x}})$ $(x_i - {\overline{x}})^2$ $8$ years $-4.4$ years $19.36$ years $^2$ $11$ years $-1.4$ years $1.96$ years $^2$ $17$ years $4.6$ years $21.16$ years $^2$ $7$ years $-5.4$ years $29.16$ years $^2$ $19$ years $6.6$ years $43.56$ years $^2$ Normally we can find the variance $({s^2})$ by averaging the squared deviations from the mean. But remember we don't know the real population mean—we had to estimate it by using the sample mean. The age of any particular zebra in our sample is likely to be closer to the average age of the 5 zebras we sampled. This is compared to the average age of all $42$ zebras in the zoo. Because of that, the squared deviations from the mean we calculated will probably underestimate the actual deviations from the population mean. To compensate for this underestimation, rather than simply averaging the squared deviations from the mean, we total them and divide by $n - 1$. $ {s^2} = \dfrac{\sum\limits_{i=1}^{{n}} (x_i - {\overline{x}})^2}{{n - 1}} $ $ {s^2} = \dfrac{{19.36} + {1.96} + {21.16} + {29.16} + {43.56}} {{5 - 1}} $ $ {s^2} = \dfrac{{115.2}}{{4}} = {28.8\text{ years}^2} $ The sample standard deviation $({s})$ is found by taking the square root of the sample variance $({s^2})$. ${s} = \sqrt{{s^2}}$ $ {s} = \sqrt{{28.8\text{ years}^2}} = {5.4\text{ years}} $ We can estimate that the average zebra at the zoo is $12.4$ years old. There is also a standard deviation of $5.4$ years.